College

A 60.0 kg object is moving east at [tex]$8.00 \, \text{m/s}$[/tex], and then slows down to [tex]$4.00 \, \text{m/s}$[/tex]. How much work was done?

A. [tex]\(-1,440 \, \text{J}\)[/tex]
B. [tex]\(-480 \, \text{J}\)[/tex]
C. [tex]\(1,440 \, \text{J}\)[/tex]
D. [tex]\(2,880 \, \text{J}\)[/tex]

Answer :

To find out how much work was done on the object, we will calculate the change in its kinetic energy.

The formula for kinetic energy (KE) is:

[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

where:
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( v \)[/tex] is the velocity of the object.

First, we calculate the initial kinetic energy when the object is moving at 8.00 m/s:

1. Initial kinetic energy:

[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 60.0 \, \text{kg} \times (8.00 \, \text{m/s})^2 \][/tex]

[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 60.0 \times 64 \][/tex]

[tex]\[ KE_{\text{initial}} = 1920.0 \, \text{Joules} \][/tex]

2. Final kinetic energy:

Next, calculate the kinetic energy after the object slows down to 4.00 m/s:

[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 60.0 \, \text{kg} \times (4.00 \, \text{m/s})^2 \][/tex]

[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 60.0 \times 16 \][/tex]

[tex]\[ KE_{\text{final}} = 480.0 \, \text{Joules} \][/tex]

3. Work done:

The work done on the object is the change in kinetic energy:

[tex]\[ \text{Work} = KE_{\text{final}} - KE_{\text{initial}} \][/tex]

[tex]\[ \text{Work} = 480.0 \, \text{J} - 1920.0 \, \text{J} \][/tex]

[tex]\[ \text{Work} = -1440.0 \, \text{Joules} \][/tex]

Therefore, the work done on the object is [tex]\(-1440 \, \text{J}\)[/tex], which indicates that energy was taken out of the system, slowing the object down. The correct choice is [tex]\(-1,440 \, \text{J}\)[/tex].